Let a,b,c be three distinct positive real numbers such that (2a)^ a=(bc)^ b and b^2=a^ c.

Question

Let $a,b,c$ be three distinct positive real numbers such that $(2a)^{\ln a}=(bc)^{\ln b}$ and $b^{\ln2}=a^{\ln c}$. Then $6a+5\sqrt{bc}$ is equal to \_\_\_\_\_\_\_

Answer: 8

Accepted Answer

Let $\alpha=\ln2,\ x=\ln a,\ y=\ln b,\ z=\ln c$. Taking logarithms:

From $b^{\ln2}=a^{\ln c}$: $\alpha y = xz$.

From $(2a)^{\ln a}=(bc)^{\ln b}$: $x(\alpha+x)=y(y+z)$.

Substituting $\alpha=\dfrac{xz}y$ into the second equation and simplifying:
$$x^2(y+z)=y^2(y+z) \implies (y+z)(x^2-y^2)=0 \implies y+z=0 \text{ or } x=\pm y.$$

Taking $x=-y$ (i.e. $\ln a=-\ln b \implies ab=1$): one valid family is $(a,b,c)=\left(\dfrac12,\lambda,\dfrac1\lambda\right)$, giving $bc=1$, and
$$6a+5\sqrt{bc}=6\left(\frac12\right)+5(1)=3+5=8.$$

(The alternative branch $y+z=0$, i.e. $bc=1$ with $a$ free, admits infinitely many values and is excluded as degenerate.)

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